Asked by clouds4587
1. Identify an equation in standard form for a hyperbola with center (0, 0), vertex (0, 13), and focus (0, 25).
2. Identify an equation in standard form for an ellipse with its center at the origin, a vertex at (0, 7), and a focus at (0, 3).
help would be much appreciated
2. Identify an equation in standard form for an ellipse with its center at the origin, a vertex at (0, 7), and a focus at (0, 3).
help would be much appreciated
Answers
Answered by
oobleck
#1. You have a vertical axis (both vertex and focus lie on the line x=0), and
a = 13
c = 25
so b^2 = c^2 - a^2 = 456
y^2/169 - x^2/456 = 1
#1. again, a vertical major axis, and
a = 7
c = 3
b^2 = a^2 - c^2 = 40
x^2/40 + y^2/49 = 1
a = 13
c = 25
so b^2 = c^2 - a^2 = 456
y^2/169 - x^2/456 = 1
#1. again, a vertical major axis, and
a = 7
c = 3
b^2 = a^2 - c^2 = 40
x^2/40 + y^2/49 = 1
Answered by
clouds4587
Thank you sir! appreciate itπ
Answered by
JANE
1.The equation of a hyperbola in standard form is given by;
(y^2/a^2)-(x^2/b^2)=1
where a and b are greater than 0 and c^2=a^2+b^2
Given vertex(0,13) and focus(0,25), then a=13 and c=25
25^2=13^2+b^2
b^2=25^2-13^2
b^2=456
The equation will therefore be;
(y^2/169)-(x^2/456)=1
(y^2/a^2)-(x^2/b^2)=1
where a and b are greater than 0 and c^2=a^2+b^2
Given vertex(0,13) and focus(0,25), then a=13 and c=25
25^2=13^2+b^2
b^2=25^2-13^2
b^2=456
The equation will therefore be;
(y^2/169)-(x^2/456)=1
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